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NDA 2020 Maths · Mathematics · Differential Equations · medium

Solving (1+x^2 dy=(1+y^2 dx by separating the variables gives the general solution

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Solving (1+x2) dy=(1+y2) dx(1+x^2)\,dy=(1+y^2)\,dx by separating the variables gives the general solution
  1. A. tan⁡−1y+tan⁡−1x=C\tan^{-1}y+\tan^{-1}x=C
  2. B. y=tan⁡x+Cy=\tan x+C
  3. C. ln⁡(1+y2)=ln⁡(1+x2)+C\ln(1+y^2)=\ln(1+x^2)+C
  4. D. tan⁡−1y−tan⁡−1x=C\tan^{-1}y-\tan^{-1}x=C
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Answer: (D) tan⁡−1y−tan⁡−1x=C\tan^{-1}y-\tan^{-1}x=C

Divide by (1+x2)(1+y2)(1+x^2)(1+y^2) to get dy1+y2=dx1+x2\dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}. Integrating both sides, tan⁡−1y=tan⁡−1x+C\tan^{-1}y=\tan^{-1}x+C, i.e. tan⁡−1y−tan⁡−1x=C\tan^{-1}y-\tan^{-1}x=C. The option with ++ has the wrong sign after moving the term, and y=tan⁡x+Cy=\tan x+C would be the solution of dydx=1+y2\dfrac{dy}{dx}=1+y^2 only for a different equation. The logarithmic form would come from integrating 2y1+y2\dfrac{2y}{1+y^2}, which is not what appears.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake